\newproblem{lay:1_9_33}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.9.33}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation. Then, there exists a unique matrix $A$ such that
	\begin{center}
		$T(\mathbf{x})=A\mathbf{x}$
	\end{center}
	In fact,
	\begin{center}
		$A=\begin{pmatrix}T(\mathbf{e}_1) & T(\mathbf{e}_2) & ... & T(\mathbf{e}_n) \end{pmatrix}$
	\end{center}
	where $\mathbf{e}_i$ is the $i$-th column of the $n\times n$ identity matrix. Show that $A$ is unique.
}{
  % Solution
	Let us assume $A$ is not unique. That is, there exists another matrix $A'\neq A$ such that $\forall \mathbf{x}\in\mathbb{R}^n$
	\begin{center}
		$T(\mathbf{x})=A'\mathbf{x}$
	\end{center}
	If we now subtract the two equations ($T(\mathbf{x})=A\mathbf{x}$ and $T(\mathbf{x})=A'\mathbf{x}$), we have
	\begin{center}
		$T(\mathbf{x})-T(\mathbf{x})=A\mathbf{x}-A'\mathbf{x}$ \\
		$\mathbf{0}=(A-A')\mathbf{x}$ \\
	\end{center}
	If this is true for all $\mathbf{x}$ is because $A-A'=0$, or what is the same, $A=A'$. But this is a contradiction with our hypothesis that $A\neq A'$ and,
	consequently $A$ is unique.
}
\useproblem{lay:1_9_33}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
